Chapter 16, Problem 36P

To determine

To find: The distance from the charge ${\mathrm{Q}}_1$ to the field point

Answer to Problem 36P

The distance from the charge ${\mathrm{Q}}_1$ to the field point is $29\text{cm}$ .

Explanation of Solution

Formula used: $\mathrm{E}=\frac{\mathrm{K}\mathrm{q}}{{\mathrm{r}}^2}$

Calculation:

The electric field due to charge ${\mathrm{Q}}_1$ at point $\mathrm{P}$ is,

  ${\mathrm{E}}_{\mathrm{A}}=\frac{\mathrm{k}{\mathrm{Q}}_1}{{\mathrm{x}}^2}$

The electric field due to charge ${\mathrm{Q}}_2$ at point $\mathrm{P}$ is,

  ${\mathrm{E}}_{\mathrm{B}}=\frac{\mathrm{k}{\mathrm{Q}}_2}{{(\mathrm{d}+\mathrm{x})}^2}$

The electric fields at P due to the two charges are equal in magnitude but opposite in direction,

because the resultant electric field is zero at a point P. Therefore,

  $\left|{\mathrm{E}}_{\mathrm{A}}\right|=\left|{\mathrm{E}}_{\mathrm{B}}\right|$

Substitute $\frac{\mathrm{k}{\mathrm{Q}}_1}{{\mathrm{x}}^2}$ for ${\mathrm{E}}_{\mathrm{A}}$ and $\frac{\mathrm{k}{\mathrm{Q}}_2}{{(\mathrm{d}+\mathrm{x})}^2}$ for ${\mathrm{E}}_{\mathrm{B}}$

  $\begin{array}{l}\frac{\mathrm{k}{\mathrm{Q}}_1}{{\mathrm{x}}^2}=\frac{\mathrm{k}{\mathrm{Q}}_2}{{(\mathrm{d}+\mathrm{x})}^2}\\ \frac{{\mathrm{Q}}_1}{{\mathrm{x}}^2}=\frac{{\mathrm{Q}}_2}{{(\mathrm{d}+\mathrm{x})}^2}\end{array}$

Rewrite the equation $\frac{{\mathrm{Q}}_1}{{\mathrm{x}}^2}=\frac{{\mathrm{Q}}_2}{{(\mathrm{d}+\mathrm{x})}^2}$ for $\mathrm{x}$

  $\begin{array}{l}{\left(\frac{\mathrm{d}+\mathrm{x}}{\mathrm{x}}\right)}^2=\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}\mathrm{d}+\mathrm{x}\\ =\mathrm{x}\left(\sqrt{\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}}\right)\mathrm{d}\\ =\mathrm{x}\left[\sqrt{\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}}-1\right]\mathrm{x}\\ =\frac{\mathrm{d}}{\sqrt{\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}}-1}\end{array}$

Substitute $25.0\mathrm{\mu }\text{C}$ for $\left|{\mathrm{Q}}_1\right|,50.0\mathrm{\mu }\text{C}$ for ${\mathrm{Q}}_2,$ and $12\text{cm}$ for $\mathrm{d}$

  $\begin{array}{l}\mathrm{x}=\frac{12\text{cm}}{\sqrt{\frac{50.0\mathrm{\mu }\text{C}}{25.0\mathrm{\mu }\text{C}}-1}}\\ =28.97\text{cm}\end{array}$

Round off to two significant figures, the distance from the charge ${\mathrm{Q}}_1$ to the field point is $29\text{cm}$ .